What is the integral of-1/(1+2x) ?

The integral of-1/(1+2x) is -1/2 ln|1+2x|+C

integral of-1/(1+2x)

\int - \frac{1}{1 + 2 x} d x

- \frac{1}{2} ln ∣ 1 + 2 x ∣ + C

Solution steps

\int - \frac{1}{1 + 2 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{1 + 2 x} d x

Apply u-substitution

= - \int \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 + 2 x

= - \frac{1}{2} ln ∣ 1 + 2 x ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ 1 + 2 x ∣ + C

\int_{1}^{8} 1 + 3 x d x

\int (x^{2} - 3 x + 6) d x

\frac{d}{d x} (x^{2} - 3 x - 4)

\frac{d}{d x} (A e^{3 x} + 3 A x e^{3 x})

\frac{\partial}{\partial x} (x cos (y - x))